Answer:
- 0.07 °C
Explanation:
At constant pressure and number of moles, Using Charle's law
Given ,
V₁ = 439 mL = 0.439 L ( 1 L = 0.001 mL )
V₂ = 0.378 L
T₁ = 317.15 K
T₂ = ?
Using above equation as:
The conversion of T(K) to T( °C) is shown below:
T( °C) = T(K) - 273.15
So, <u>T = 273.08 - 273.15 °C = - 0.07 °C</u>
Answer:
142.36 kg
Explanation:
volume of water in the lake = surface area x depth
= 18.5 x ( 1760 x 3 )² x 39 ft³
= 2.011 x 10¹⁰ ft³
= 2.011 x 10¹⁰ x 28.3168 liter .
= 56.945 x 10¹⁰ liter .
concentration of mercury = .25 x 10⁻⁶ g / liter
= 25 x 10⁻⁸ g / liter
= 25 x 10⁻¹¹ kg / liter
mass of mercury in the water of lake
= 25 x 10⁻¹¹ x 56.945 x 10¹⁰ kg
= 142.36 kg .
The final concentration of the diluted standard is 0.2 mg/dL.
<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>
Using the dilution formula:
where
- C1 is initial concentration
- V1 initial volume
- C2 is final concentration
- V2 is final volume.
Assuming a final volume of 100 mL, and since a 1/5 dilution is made:
C1 = 1.00 mg/dL
V1 = 20
C2 = ?
V2 = 100 mL
C2 = C1V1/V2
C2 = 20 × 1/100
C2 = 0.2 mg/dL
Therefore, the final concentration of the diluted standard is 0.2 mg/dL.
Learn more about dilution at: brainly.com/question/24881505